Simple Nature - Light and Matter

Problem 43.43 On pp. 884-885 of subsection 13.4.4, we used simple algebra toderive an approximate expression for the energies of states in hydrogen,without having to explicitly solve the Schrödinger equation. Asinput to the calculation, we used the the proportionality U ∝ r −1 ,which is a characteristic of the electrical interaction. The result forthe energy of the nth standing wave pattern was E n ∝ n −2 .There are other systems of physical interest in which we haveU ∝ r k for values of k besides −1. Problem 23 discusses the groundstate of the harmonic oscillator, with k = 2 (and a positive constantof proportionality). In particle physics, systems called charmoniumand bottomonium are made out of pairs of subatomic particlescalled quarks, which interact according to k = 1, i.e., a forcethat is independent of distance. (Here we have a positive constantof proportionality, and r > 0 by definition. The motion turns outnot to be too relativistic, so the Schrödinger equation is a reasonableapproximation.) The figure shows actual energy levels for thesethree systems, drawn with different energy scales so that they canall be shown side by side. The sequence of energies in hydrogenapproaches a limit, which is the energy required to ionize the atom.In charmonium, only the first three levels are known. 8Generalize the method used for k = −1 to any value of k, andfind the exponent j in the resulting proportionality E n ∝ n j . Comparethe theoretical calculation with the behavior of the actual energiesshown in the figure. Comment on the limit k →√∞.44 The electron, proton, and neutron were discovered, respectively,in 1897, 1919, and 1932. The neutron was late to the party,and some physicists felt that it was unnecessary to consider it asfundamental. Maybe it could be explained as simply a proton withan electron trapped inside it. The charges would cancel out, givingthe composite particle the correct neutral charge, and the massesat least approximately made sense (a neutron is heavier than a proton).(a) Given that the diameter of a proton is on the order of10 −15 m, use the Heisenberg uncertainty principle to estimate the√trapped electron’s minimum momentum.√(b) Find the electron’s minimum kinetic energy.(c) Show via E = mc 2 that the proposed explanation fails, becausethe contribution to the neutron’s mass from the electron’s kineticenergy would be many orders of magnitude too large.8 See Barnes et al., “The XYZs of Charmonium at BES,” arxiv.org/abs/hep-ph/0608103. To avoid complication, the levels shown are only those in thegroup known for historical reasons as the Ψ and J/Ψ.902 Chapter 13 Quantum Physics